How do I get the filepath of an uploaded file?

Question

I'm having difficulty passing the path of a file to a library called Textract.

def file_to_string(filepath):
    text = textract.process(filepath)
    print text
    return text

Here is my upload form in views.py

if request.method == 'POST':
    upload_form = UploadFileForm(request.POST, request.FILES)
    if upload_form.is_valid():  
        file = request.FILES['file']
        filetosave = File(file=file, filename=file.name)
        filetosave.save()
        if validate_file_extension(file):
            request.session['text'] = file_to_string(file) # something in here
else:
    upload_form=UploadFileForm()

models.py

class File(models.Model):
    filename = models.CharField(max_length=200)
    file = models.FileField(upload_to='files/%Y/%m/%d')
    upload_date=models.DateTimeField(auto_now_add =True)
    status = models.CharField(max_length=200)

    def __unicode__(self):
        return self.filename

Now Textract expects a path to go into file_to_string(filepath). If I try to pass in the file object it gives me an error: "coercing to Unicode: need string or buffer, InMemoryUploadedFile found".

But if it is an InMemoryUploadedFile type, how do I get the path? I understand this is stored in memory and doesn't have a path.

How should I handle this -- should I save the file object first and then try to access it? If I save the file and then try request.session['text'] = file_to_string(file.name) it gives a MissingFileError, though the docs say that this should give the name of the file including the relative path from MEDIA_ROOT.

Thanks a lot in advance.

Answer 1

request.session['text'] = file_to_string(filetosave.file.path) should do the trick